document.write( "Question 79742: What is an extraneous solutions? \n" ); document.write( "
Algebra.Com's Answer #57215 by tutor_paul(519)\"\" \"About 
You can put this solution on YOUR website!
The best way to understand what extraneous roots are is to go through a problem where you get one. The following is a solution to a problem that I did previously which dealt with extraneous roots:\r
\n" ); document.write( "\n" ); document.write( "\"sqrt%282y%2B7%29%2B4=y\"
\n" ); document.write( "First, get the radical expression alone on one side of the equation:
\n" ); document.write( "\"sqrt%282y%2B7%29=y-4\"
\n" ); document.write( "Now, square both sides of the equation to get rid of the radical:
\n" ); document.write( "\"2y%2B7=%28y-4%29%5E2\"
\n" ); document.write( "Simplify the right hand side:
\n" ); document.write( "\"2y%2B7=y%5E2-8y%2B16\"
\n" ); document.write( "Combine like terms and equate to zero:
\n" ); document.write( "\"y%5E2-10y%2B9=0\"
\n" ); document.write( "Factor the expression:
\n" ); document.write( "\"%28y-9%29%28y-1%29=0\"
\n" ); document.write( "Equate each factor to zero, and solve:
\n" ); document.write( "\"y=9\" and \"y=1\"\r
\n" ); document.write( "\n" ); document.write( "Now, this part is important...you need to plug these answers back in to
\n" ); document.write( "the original equation to be sure they are not \"extraneous.\" An extraneous
\n" ); document.write( "root may be mathematically correct, but it is not the true answer. If you
\n" ); document.write( "plug y=1 back into the original equation, you will see that the equation
\n" ); document.write( "DOES NOT hold true. Hence, this is an extraneous root. If you plug y=9 back
\n" ); document.write( "into the original equation, you will see that the equation DOES hold true,
\n" ); document.write( "so that one is your answer.\r
\n" ); document.write( "\n" ); document.write( "Good Luck,
\n" ); document.write( "tutor_paul@yahoo.com
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